Varsity Math, Week 18

Column number

1

2

3

4

5

A

0

?

X

0

0

C

X

0

0

?

?

E

0

1

?

?

2

R

?

2

?

0

?

T

?

?

1

1

2

Much to the delight of the members of the Varsity Math team, it can happen that some word puzzles are math puzzles, too!

Word Block

Casey has a five-by-five grid with one of the letters A, C, E, R or T in each space. All five letters appear in each row of the grid. The number of occurrences of each letter in each column of the grid is given by the table at right (with some missing entries, to be determined) in which X represents the square root of the answer to Staircase Surface from last week. Moreover, the two vowels are adjacent in at most one row.

What are the five rows of Casey’s grid?

[asciimathsf]

Double Veracity

The team is holding an informal practice late one afternoon when Jayden says to Amari, “I’ve made a great discovery!” and holds up a sheet of paper on which is written “SIN² + COS² = UNITY”. Amari responds, “Well, yeah, everyone on the team knows that for any angle, the sum of the squares of the sine and cosine of that angle is one. Big deal!” “But that’s not all,” says Jayden, “you can also replace each letter in that expression with a digit so that the same letter is always replaced by the same digit and different letters by different digits, so that it remains a true equation.” And then after a minute Jayden adds, “You may not really need to know this, but I think it’s cool that with the same assignments to letters in common and the same rules, SINE² + COSINE² also has a ONE in it, because it comes out to SESBYEONEEE.”

What digit should replace each letter in the first equation? (Note that although there is already a “2” in the exponents, one of the letters is allowed to correspond to the digit 2 if need be.)

Solutions to Week 17

Beaver Bash. If you play a few rounds of the game, you will discover that it seems as though the two options for the last player always lead to one configuration with an odd number of loops and one with an even number of loops. To see if we can figure out why that would be so, imagine the section of the board surrounding the last unused turntable, which is shown blank in the accompanying diagram. The four “loose ends” of the tracks that have been established are labeled N, E, S, W after the compass points. Imagine a beaver starting at the point marked “N” and heading north. It will wend its way around various tracks in the diagram, but eventually it must reach one of the other three loose ends, E, S or W, because all of the other tracks are set and match up — there are no other dead ends it could get to besides E, S or W, and there’s no way it could make a U-turn and come back to N. Similarly, the remaining two loose ends must connect to each other, since they can’t somehow merge with the track reached from the point N. But now we can deduce that N can’t connect to S, for if it did, the track running from E to W would have to cross it somewhere else in the grid, and the exhibit is constructed so that tracks can never cross. Hence, either N connects to E and S connects to W, or else N connects to W and S connects to E. In either case, in one position for the last turntable, these two sections of track will be joined into one big loop, and in the other position, they will each make their own loop. For example, if N connects to E and S connects to W, then when the last turntable is set so that it also connects N to E and S to W, then there will be two loops, and in the other position the two track sections will be linked into a single big loop.

Since 1 is an odd number and 2 is an even number, by choosing the orientation of that last turntable, the last player can determine whether the total number of loops is odd or even. Therefore, the last player wins; and since there are an even number of turntables, the winner/last player will always be the second player.